Contents 1 Hypotheses 2 Case of a cylinder-cylinder contact 2.1 Negligible clearance and rigid bodies 2.2 Negligible clearance and elastic bodies 2.3 Clearance and elastic bodies 3 Case of a sphere-sphere contact 3.1 Case of uniform pressure 3.2 Case of a sinusoidal repartition of pressure 4 Hertz contact stress 5 Applications 5.1 Bolt used as a stop 5.2 Plain bearing 6 References 7 Bibliography Hypotheses[edit] A contact between a male part (convex) and a female part (concave) is considered when the radii of curvature are close to one another. There is no tightening and the joint slides with no friction therefore, the contact forces are normal to the tangent of the contact surface. Moreover, bearing pressure is restricted to the case where the charge can be described by a radial force pointing towards the center of the joint. Case of a cylinder-cylinder contact[edit]
In the case of a revolute joint or of a hinge joint, there is a contact between a male cylinder and a female cylinder. The complexity depends on the situation, and three cases are distinguished: the clearance is negligible: a) the parts are rigid bodies, b) the parts are elastic bodies; c) the clearance cannot be ignored and the parts are elastic bodies. By "negligible clearance", H7/g6 fit is typically meant. The axes of the cylinders are along the z-axis, and two external forces apply to the male cylinder: a force F → displaystyle vec F along the y-axis, the load; the action of the bore (contact pressure). The main concern is the contact pressure with the bore, which is uniformly distributed along the z-axis. Notation: D is the nominal diameter of both male and female cylinders;[2] L the guiding length. Negligible clearance and rigid bodies[edit] Uniform bearing pressure: case of rigid bodies when the clearing can be neglected. In this first modeling, the pressure is uniform. It is equal to[3] · [4] · :[5] P = F D × L = r a d i a l l o a d p r o j e c t e d a r e a displaystyle P= frac F Dtimes L = frac mathrm radial load mathrm projected area . Proof There are two ways to obtain this result. Hemicylindrical body at the equilibrium in a fluid with hydrostatic pressure. First, we can consider a hemicylinder in a fluid, with a uniform hydrostatic pressure. The equilibrium is achieved when the resulting force on the flat surface is equal to the resulting force on the curved one. The flat surface is a D × L rectangle, therefore F = P × (D × L) q.e.d. The elementary force dF, due to the pressure on a surface element dS, has two components: dFx and dFy. Second, we can integrate the pressure elementary forces. Let us consider a small surface dS on the cylindrical part, parallel to a generating line; its length is L, and it is bound by the angles θ and θ + dθ. This small surface element can be considered as a flat rectangle which dimensions are L × (dθ × D/2). The pressure force on the surface is equal to dF = P × dS = ½ × P × D × L × dθ The (y, z) plane is a plane of reflection symmetry, so the x compound of this force is annihilated by the force on the symmetrical surface element. The y compound of this force is equal to: dFy = cos(θ) dF = ½ × cos(θ) × P × D × L × dθ. The resulting force is equal to F y = ∫ − π / 2 π / 2 1 2 × P × D × L × cos ( θ ) × d θ = 1 2 × P × D × L × [ sin ( θ ) ] − π / 2 π / 2 = P × D × L displaystyle F_ y =int _ -pi /2 ^ pi /2 frac 1 2 times Ptimes Dtimes Ltimes cos(theta )times mathrm d theta = frac 1 2 times Ptimes Dtimes Ltimes left[sin(theta )right]_ -pi /2 ^ pi /2 =Ptimes Dtimes L q.e.d. This calculation is similar to the case of a cylindrical vessel under pressure. Negligible clearance and elastic bodies[edit]
If it is considered that the parts deform elastically, then the contact pressure is no longer uniform and transforms to a sinusoidal repartition[6] · :[7] P(θ) = Pmax⋅cos θ with P m a x = 4 π ⋅ F L D displaystyle P_ mathrm max = frac 4 pi cdot frac F LD . This is a particular case of the following section (θ0 = π/2). The maximum pressure is 4/π ≃ 1.27 times bigger than the case of uniform pressure. Clearance and elastic bodies[edit]
In cases where the clearance can not be neglected, the contact between the male part is no longer the whole half-cylinder surface but is limited to a 2θ0 angle. The pressure follows Hooke's law:[8] P(θ) = K⋅δα(θ) where K is a positive real number that represents the rigidity of the materials; δ(θ) is the radial displacement of the contact point at the angle θ; α is a coefficient that represents the behaviour of the material: α = 1 for metals (purely elastic behaviour), α > 1 for polymers (viscoelastic or viscoplastic behaviour). The pressure varies as: A⋅cos θ - B where A and B are positive real number. The maximum pressure is: P m a x = 4 F L D × 1 − cos θ 0 2 θ 0 − sin 2 θ 0 displaystyle P_ mathrm max = frac 4F LD times frac 1-cos theta _ 0 2theta _ 0 -sin 2theta _ 0 the angle θ0 is in radians. The rigidity coefficient K and the half contact angle θ0 can not be derived from the theory. They must be measured. For a given system — given diameters and materials —, thus for given K and clearance j values, it is possible to obtain a curve θ0 = ƒ(F/(DL)). Proof Elastic deformation in case of a male-female cylinders contact. Relationship between pressure, clearance and contact angle The part no. 1 is the containing cylinder (female, concave), the part no. 2 is the contained cylinder (male, convex); the center of the cylinder i is Oi, and its radius is Ri. The reference position is an ideal situation where both cylinders are concentric. The clearing, expressed as a radius (not diameter), is: j = R1 - R2. Under the load, the part 2 gets in contact with the part 1, the he surfaces deform. we suppose that the cylinder 2 is rigid (no deformation), and that the cylinder 1 is an elastic body. The indentation of 2 into 1 has a depth of δmax; the cylinder movement is e (excentration): e = O1O2 = j + δmax. We considere the frame at the center of the cylinder 1 (O1, x, y). Let M be a point on the contact surface; θ is the angle (-y, O1M). The displacement of the surface, δ, is: δ(θ) = O1M - R1. with δ(0) = δmax. The coordinates of M are: M((R1 + δ(θ)⋅sin θ) ; -(R1 + δ(θ))⋅cos θ) and the coordinates of O2: O2(0 ; -e). Let us considere the frame (O1, u, v), where the axis u is (O1M). In this frame, the coordinates are: M(R1 + δ(θ) ; 0) O2(e⋅cos θ ; -e⋅sin θ) We know that O 1 M → = O 1 O 2 → + O 2 M → displaystyle overrightarrow O_ 1 M = overrightarrow O_ 1 O_ 2 + overrightarrow O_ 2 M thus R 1 + δ ( θ ) = e ⋅ cos θ + R 2 cos φ 0 = − e ⋅ sin θ + R 2 sin φ displaystyle left begin aligned R_ 1 +delta (theta )&=&ecdot cos theta +R_ 2 cos varphi \0&=&-ecdot sin theta +R_ 2 sin varphi end aligned right. then we use the expression of e and R1 = j + R2: j + R 2 + δ ( θ ) = ( j + δ max ) ⋅ cos θ + R 2 cos φ 0 = − ( j + δ max ) ⋅ sin θ + R 2 sin φ displaystyle left begin aligned j+R_ 2 +delta (theta )&=&(j+delta _ max )cdot cos theta +R_ 2 cos varphi \0&=&-(j+delta _ max )cdot sin theta +R_ 2 sin varphi end aligned right. The deformations are small, as we are in the elastic domain. Thus, δmax ≪ R1 and therefore φ ≪ 1, i.e. cos φ ≃ 1 sin φ ≃ 1 (in radians) thus ( j + δ max ) ⋅ cos θ + R 2 − ( j + R 2 + δ ( θ ) ) = 0 ( j + δ max ) ⋅ sin θ − R 2 ⋅ φ = 0 displaystyle left begin aligned (j+delta _ max )cdot cos theta +R_ 2 -(j+R_ 2 +delta (theta ))&=&0\(j+delta _ max )cdot sin theta -R_ 2 cdot varphi &=&0end aligned right. and ( j + δ max ) ⋅ cos θ − j − δ ( θ ) = 0 ( j + δ max ) ⋅ sin θ − R 2 ⋅ φ = 0 displaystyle left begin aligned (j+delta _ max )cdot cos theta -j-delta (theta )&=&0\(j+delta _ max )cdot sin theta -R_ 2 cdot varphi &=&0end aligned right. At θ = θ0, δ(0) = 0 and the first equation is ( j + δ max ) ⋅ cos θ 0 − j = 0 ⇒ δ max = j ( 1 − cos θ 0 ) cos θ 0 ⇔ cos θ 0 = j j + δ max displaystyle (j+delta _ max )cdot cos theta _ 0 -j=0Rightarrow delta _ max = frac j(1-cos theta _ 0 ) cos theta _ 0 Leftrightarrow cos theta _ 0 = frac j j+delta _ max and thus ( j + j ( 1 − cos θ 0 ) cos θ 0 ) ⋅ cos θ − j − δ ( θ ) = 0 displaystyle left(j+ frac j(1-cos theta _ 0 ) cos theta _ 0 right)cdot cos theta -j-delta (theta )=0 ⟹ δ ( θ ) = j ( cos θ cos θ 0 − 1 ) displaystyle Longrightarrow delta (theta )=jleft( frac cos theta cos theta _ 0 -1right) [1]. If we use the law of elasticity for a metal (α = 1): P ( θ ) = K ⋅ j ( cos θ cos θ 0 − 1 ) P max = K ⋅ j ⋅ 1 − cos θ 0 cos θ 0 displaystyle left begin aligned P(theta )&=&Kcdot jleft( frac cos theta cos theta _ 0 -1right)\P_ max &=&Kcdot jcdot frac 1-cos theta _ 0 cos theta _ 0 end aligned right. [2] The pressure is an affine function of cos θ: P(θ) = A⋅cos θ - B with A = K⋅j/cos θ0 and B = A⋅cos θ0. Case where the clearance can be neglected If j ≃ 0 (R1 ≃ R2), then the contact is on the whole half-perimeter: 2θ0 ≃ π and cos θ0 ≃ 0. The value of 1/cos θ0 rise towards infinity, thus δ ( θ ) ≃ j cos θ cos θ 0 displaystyle delta (theta )simeq j frac cos theta cos theta _ 0 As j and cos θ0 both tend towards 0, the ratio j/cos θ0 is not defined when j goes to 0. In mechanical engineering, j = 0 is an uncertain fit, it is a nonsense, both mathematically and mechanically. We are looking for a limit function P π / 2 ( θ ) = lim θ 0 → π / 2 P θ 0 ( θ ) displaystyle P_ pi /2 (theta )=lim _ theta _ 0 to pi /2 P_ theta _ 0 (theta ) . So, the pressure is a sinusoid function of θ: P ( θ ) = K ⋅ j cos θ 0 ⋅ cos θ displaystyle P(theta )=Kcdot frac j cos theta _ 0 cdot cos theta thus P(θ) = Pmax⋅cos θ with P max = K ⋅ j cos θ 0 displaystyle P_ max =Kcdot frac j cos theta _ 0 . Let us considere an infinitesimal element of surface dS bound by θ and θ + dθ. As in the case of the uniform pressure, we have dFy(θ) = cos(θ)dF = ½ × cos(θ) × P(θ) × D × L × dθ = ½ × cos2(θ) × Pmax × D × L × dθ. When we integrate between -π/2 and π/2, the result is: F = ∫ − π / 2 π / 2 d F y ( θ ) d θ = 1 2 P max D L ∫ − π / 2 π / 2 cos 2 θ d θ displaystyle F=int _ -pi /2 ^ pi /2 mathrm d F_ y (theta )mathrm d theta = frac 1 2 P_ max DLint _ -pi /2 ^ pi /2 cos ^ 2 theta mathrm d theta We know that (e.g. using the Euler's formula): ∫ − π / 2 π / 2 cos 2 θ d θ = 1 4 [ 2 θ + sin 2 θ ] − π / 2 π / 2 = 1 2 [ θ + sin θ cos θ ] − π / 2 π / 2 = π 2 displaystyle int _ -pi /2 ^ pi /2 cos ^ 2 theta mathrm d theta = frac 1 4 left[2theta +sin 2theta right]_ -pi /2 ^ pi /2 = frac 1 2 left[theta +sin theta cos theta right]_ -pi /2 ^ pi /2 = frac pi 2 therefore F = π 4 P max D L displaystyle F= frac pi 4 P_ max DL and thus P max = 4 π F D L displaystyle P_ max = frac 4 pi frac F DL q.e.d. Case where the clearance can not be neglected The force on an infinitesimal element of surface is: dF(θ) = P(θ)dS = Kδ(θ)dS = ½ × K × j × cos θ/cos θ0 - 1) × dS d F y = K j 2 ( cos θ cos θ 0 − 1 ) D L cos θ d θ displaystyle mathrm d F_ y = frac Kj 2 left( frac cos theta cos theta _ 0 -1right)DLcos theta mathrm d theta F = K j D L 2 ∫ − θ 0 θ 0 ( cos 2 θ cos θ 0 − cos θ ) d θ = K j D L 2 [ θ + sin θ cos θ 2 cos θ 0 − sin θ ] − θ 0 θ 0 displaystyle F= frac KjDL 2 int _ -theta _ 0 ^ theta _ 0 left( frac cos ^ 2 theta cos theta _ 0 -cos theta right)mathrm d theta = frac KjDL 2 left[ frac theta +sin theta cos theta 2cos theta _ 0 -sin theta right]_ -theta _ 0 ^ theta _ 0 thus F = K j D L 2 ( 2 θ 0 + 2 sin θ 0 cos θ 0 2 cos θ 0 − 2 sin θ 0 ) = K j D L 2 ( θ 0 − sin θ 0 cos θ 0 cos θ 0 ) displaystyle F= frac KjDL 2 left( frac 2theta _ 0 +2sin theta _ 0 cos theta _ 0 2cos theta _ 0 -2sin theta _ 0 right)= frac KjDL 2 left( frac theta _ 0 -sin theta _ 0 cos theta _ 0 cos theta _ 0 right) . We recognise the trigonometric identity sin 2θ = 2 sin θ cos θ : F = K j D L 4 cos θ 0 ( 2 θ 0 − sin 2 θ 0 ) displaystyle F= frac KjDL 4cos theta _ 0 (2theta _ 0 -sin 2theta _ 0 ) thus K = 4 F cos θ 0 j D L ( 2 θ 0 − sin 2 θ 0 ) displaystyle K= frac 4Fcos theta _ 0 jDL(2theta _ 0 -sin 2theta _ 0 ) and therefore: P max = K j 1 − cos θ 0 cos θ 0 = 4 F D L × 1 − cos θ 0 2 θ 0 − sin 2 θ 0 displaystyle P_ max =Kj frac 1-cos theta _ 0 cos theta _ 0 = frac 4F DL times frac 1-cos theta _ 0 2theta _ 0 -sin 2theta _ 0 q.e.d. Case of a sphere-sphere contact[edit]
A sphere-sphere contact corresponds to a spherical joint (socket/ball), such as a ball jointed cylinder saddle. It can also describe the situation of bearing balls. Case of uniform pressure[edit] The case is similar as above: when the parts are considered as rigid bodies and the clearance can be neglected, then the pressure is supposed to be uniform. It can also be calculated considering the projected area[3] · [9] · :[10] P = F π R 2 = r a d i a l l o a d p r o j e c t e d a r e a displaystyle P= frac F pi R^ 2 = frac mathrm radial load mathrm projected area . Case of a sinusoidal repartition of pressure[edit] As in the case of cylinder-cylinder contact, when the parts are modeled as elastic bodies with a negligible clearance, then the pressure can be modeled with a snusoidal repartition[6] · :[11] P(θ, φ) = Pmax⋅cos θ with P = 3 2 ⋅ F π R 2 displaystyle P= frac 3 2 cdot frac F pi R^ 2 . Hertz contact stress[edit]
Main article:
Hertz contact stress in the case of a male cylinder-female cylinder contact. When the clearance can not be neglected, it is then necessary to know the value of the half contact angle θ0 , which can not be determined in a simple way and must be measured. When this value is not available, the Hertz contact theory can be used. The Hertz theory is normally only valid when the surfaces can not conform, or in other terms, can not fit each other by elastic deformation; one surface must be convex, the other one must be also convex or plane. This is not the case here, so the results must be considered with great care. The approximation is only valid when the inner radius of the container R1 is far greater than the outer radius of the content R2, in which case the surface container is then seen as flat by the content. However, in all cases, the pressure that is calculated with the Hertz theory is greater than the actual pressure (because the contact surface of the model is smaller than the real contact surface), which affords designers with a safety margin for their design. In this theory, the radius of the female part (concave) is negative.[12] A relative diameter of curvature is defined: 1 d ∗ = 1 d 1 + 1 d 2 displaystyle frac 1 d^ * = frac 1 d_ 1 + frac 1 d_ 2 where d1 is the diameter of the female part (negative) and d2 is the diameter of the male part (positive). An equivalent module of elasticity is also defined: 1 E ∗ = 1 − ν 1 2 E 1 + 1 − ν 2 2 E 2 displaystyle frac 1 E^ * = frac 1-nu _ 1 ^ 2 E_ 1 + frac 1-nu _ 2 ^ 2 E_ 2 where νi is the
b = ( 2 F π L ⋅ d ∗ E ∗ ) 1 / 2 displaystyle b=left( frac 2F pi L cdot frac d^ * E^ * right)^ 1/2 and the maximal pressure is in the middle: P max = 2 F π b L = 2 F E ∗ π L d ∗ displaystyle P_ max = frac 2F pi bL = sqrt frac 2FE^ * pi Ld^ * . Hertz contact stress in the case of a male sphere-female sphere contact. In case of a sphere-sphere contact, the contact surface is a disk whose radius is: a = ( 3 F 8 ⋅ d ∗ E ∗ ) 1 / 3 displaystyle a=left( frac 3F 8 cdot frac d^ * E^ * right)^ 1/3 and the maximal pressure is in the middle: P max = 3 F π a 2 = 4 π 3 F ( E ∗ d ∗ ) 2 3 displaystyle P_ max = frac 3F pi a^ 2 = frac 4 pi sqrt[ 3 ] 3Fleft( frac E^ * d^ * right)^ 2 . Applications[edit] Bolt used as a stop[edit]
Dimensions used to design a bolted connection according to the
In a bolted connection, the role of the bolts is normally to press one parts on the other; the adherence (friction) is opposed to the tangent forces and prevents the parts from sliding apart. In some cases however, the adherence is not sufficient. The bolts then play the role of stops: the screws endure shear stress whereas the hole endure bearing pressure. In good design practice, the threaded part of the screw should be small and only the smooth part should be in contact with the plates; in the case of a shoulder screw, the clearance between the screw and the hole is very small ( a case of rigid bodies with negligible clearance). If the acceptable pressure limit Plim of the material is known, the thickness t of the part and the diameter d of the screw, then the maximum acceptable tangent force for one bolt Fb, Rd (design bearing resistance per bolt) is: Fb, Rd = Plim × d × t. In this case, the acceptable pressure limit is calculated from the
ultimate tensile stress fu and factors of safety, according to the
Plim = 1.5 × fu/γM2 where γM2 = 1.25: partial safety factor. In more complex situations, the formula is: Plim = k1 × α × fu/γM2 where k1 and α are factors that take into account other failure modes than the bearing pressure overload; k1 take into account the effects that are perpendicular to the tangent force, and α the effects along the force; k1 = min 2.8e2/d0 ; 2.5 for end bolts, k1 = min 1.4p2/d0 ; 2.5 for inner bolts, e2: edge distance from the centre of a fastener hole to the adjacent edge of the part, measured at right angles to the direction of load transfer, p2: spacing measured perpendicular to the load transfer direction between adjacent lines of fasteners, d0: diameter of the passthrough hole; α = min e1/3d0 ; p1/3d0 - 1/4 ; fub/fu ; 1 , with e1: end distance from the center of a fastener hole to the adjacent end of the part, measured in the direction of load transfer, p1: spacing between centers of fasteners in the direction of load transfer, fub: specified ultimate tensile strength of the bolt. Ultimate tensile stress for usual structural steels[1] · [13] Steel grades (EN standard) S235 S275 S355 Ultimate tensile stress fu (MPa) 360 430 510 When the parts are in wood, the acceptable limit pressure is about 4 to 8.5 MP.[14] Plain bearing[edit] In plain bearings, the shaft is usually in contact with a bushing (sleeve or flanged) to reduce friction. When the rotation is slow and the load is radial, the model of uniform pressure can be used (small deformations and clearance). The product of the bearing pressure times the circumferential sliding speed, called load factor PV, is an estimation of the resistance capacity of the material against the frictional heating[15] · [16] · .[17] Acceptable bearing pressure[18] Type of bushing Maximal circumferential sliding speed Acceptable bearing pressure (MPa) Self-lubricating bushels 7 to 8 m/s 13 m/s for graphite graphite: 5 lead bronze: 20 to 30 tin bronze: 7 to 35 Composite bushing, Glacier 2 to 3 m/s acetal: 70 PTFE: 50 Polymer bushing 2 to 3 m/s 7 to 10 References[edit] ^ a b c EN 1993-1-8:2005 Eurocode 3: Design of steel structures - Part 1-8: Design of joints ^ due to the clearance, the diameter of the bore is bigger than the diameter of the male cylinder; however, we suppose that the diameters are close to each othert ^ a b (SG 2003, p. 139) ^ (GCM 2000, p. 177) ^ (Aublin 1992, pp. 108, 136) ^ a b (SG 2003, p. 140) ^ (Aublin 1992, pp. 120–122, 136–137) ^ (Aublin 1992, pp. 120–122, 137–138) ^ (GCM 2000, pp. 110–111) ^ (Aublin 1992, pp. 108, 144–145) ^ (Aublin 1992, pp. 120–122, 145–150) ^ (Fanchon 2001, pp. 467–471) ^ a b Seinturier, Francine. "C-viii Assemblages boulonnés". Construction métallique 2 (PDF) (in French). IUT Grenoble I. ^ MB (April 2007). "Assemblages". Wiki de l'Unité Construction de Gramme (in French). Retrieved 2015-11-25. ^ (Fanchon 2011, p. 255) ^ (Chevalier 2004, p. 258) ^ (GCM 2000, pp. 113–116, 176–181) ^ L.P. Pierre et Marie Curie, Aulnoye. "Paliers lisses ou coussinets". Construction mécanique (PDF) (in French). Université de Toulon. Bibliography[edit] [Aublin 1992] Aublin, Michel; Boncompain, René; Boulaton, Michel; Caron, Daniel; Jeay, Émile; Lacage, Bernard; Réa, Jacky (1992). Systèmes mécaniques : théorie et dimensionnement (in French). Dunod. pp. 108–157. ISBN 2-10-001051-4. [Chevalier 2004] Chevalier, André (2004). Guide du dessinateur industriel (in French). Hachette technique. p. 258. ISBN 978-2-01-168831-6. [Fanchon 2001] Fanchon, Jean-Louis (2001). Guide de mécanique : sciences et technologies industrielles (in French). Nathan. pp. 467–471. ISBN 978-2-09-178965-1. [Fanchon 2011] Fanchon, Jean-Louis (2011). "Calcul des coussinets (régime non hydrodynamique)". Guide des sciences et technologies industrielles (in French). Afnor/Nathan. pp. 255–256. ISBN 978-2-09-161590-5. [GCM 2000] Texeido, C.; Jouanne, J.-C.; Bauwe, B.; Chambraud, P.; Ignatio, G.; Guérin, C. (2000). Guide de construction mécanique (in French). Delagrave. pp. 110–116, 176–180. ISBN 978-2-206-08224-0. [SG 2003] Spenlé, D.; Gourhant, R. (2003). Guide du calcul en mécanique : maîtriser la performance des systèmes industriels (in French). Hachette technique. pp. 139–140. ISBN 2-0 |